1.

Identical springs of steel and copper are equally stretched. On which more work will have to be done?

Answer»

Solution :Work done in STRETCHING a spring, `W=1/2 F xx DELTAL`
where `Delta l` increases in length DUE to applying force,
` therefore W prop Delta l[ because (1)/(2) , F` CONSTANT]
`therefore W prop (l)/(Y) [ because Delta l = (Fl )/(AY), F, l, A` are constant]
`therefore (W_("steel "))/(W_("copper")) = (Y _("copper "))/( Y _(steel"))`
but ` Y _("steel " ) lt W _("copper")`
`therefore ` Hence, more work will be done in CASE in spring made of copper.


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